Verified answer

I1 = V/R1 = 6/2 = 3.0 A

I2 = V/R2 = 6/3 = 2.0 A

I3 = V/R3 = 6/6 = 1.0A

I batt = I1+I2+I3 = 3+2+1 = 6.0 A

In a parallel circuit The current splits at each junction. The PD stays the same all across the circuit and isnt "used up" at any point. From your question I'm guessing you have a battery and the A separate resistor on one branch of the circuit. If this is the case, then you can use V=IR for each resistor individually.

For R1: R=2 and V=6 (This will always be 6 in this circuit) So Re arrange V=IR to I=V/R for current. So I=6/2 =3 Amps

R2: R=3 V=6 I=6/3 =2 Amps

R3: R=6 V=6 I=6/6=1 Amps

Method for I across the battery: V=IR for entire circuit. V=6 still. R is more complicated. To get the resistance when resistors are parallel, The formula is 1/Rtotal=1/R1+1/R2+.......

This gives 1/Rt=1/2+1/3+1/6

1/Rt=1/1

So R=1

I=V/R

I=6/1

I=6 Amps across battery

Since Voltage is equal in each resistor because its parallel, you can use Ohm's law for each one ( I=V/R)

I(in R1) =6/2 = 3A

I(in R2)=6/3=2A

I(in R3)=6/6=1A

Oh I missed the battery

First of all you need to find Req

In parallel 1/Req=1/R1+1/R2+1/R3

1/Req=1/2+1/3 +1/6= 3/6 + 2/6 +1/6 =1

Req=1

Then I=6/1=6A

You can also just add the current of each resistor since its parallel

Draw the diagram if its not with the problem. Notice that there is no way to isolate each individual resistor to get a voltage change. Since all of one resistor is hooked to the same side as the other resistors, the voltage across each resistor is the same as the voltage of the battery. To get the current through each resistor us Ohm’s Law V = R I . The current through the battery will be the sum of the currents through the resistors It = I1 + I2 + I3 ,

Resistors in series can be added as R1 + R2 + R3

For parallel circuits resistors are added as 1/R1 + 1/R2 + 1/R3

All 6 volts must be "used up" in overcoming the total resistance.

From there it's simply working out the proportion of voltage "used up" over each resistor.

You should be able to work it out from this - Hope it helps.

resistors in parallel are product over sum so 2 10 ohms become 10x10/10+10 i;e 5 ohms, when you've learnt this, you can use 1/10 + 1/10 = 0.2 1/x = 5

so 1/2+1/3+1/6 = 1 1/x = 1